Problem 157 - Algebra and Trigonometry

Question

Explain how to solve \(x^{2}+6 x+8=0\) by completing the square.

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2}+6x+8=0\) are \(x=-2\) and \(x=-4\).

Step 1: Write the equation in standard form

A quadratic equation has the standard form \(ax^{2}+bx+c=0\). Therefore, our equation, \(x^{2}+6x+8=0\) is already in standard form.

Step 2: Write the equation missing its constant term

By writing the quadratic equation without the constant term we get \(x^{2} + 6x = -8\). This is done in order to create a perfect square trinomial.

Step 3: Complete the square

Completing the square involves inserting an additional term \((b/2)^{2}\) that completes the square trinomial. The \(b\) from equation \(ax^{2}+bx+c=0\) is the coefficient of the linear \(x\), hence divide 6 by 2 and square the result to get 9 (i.e., \((3)^2 = 9\)). Therefore, add 9 to both sides to maintain the balance of the equation. This will give: \(x^{2}+6x+9=-8+9\). Simplifying the right-hand side gives \(x^{2}+6x+9=1\). Here, \(x^{2}+6x+9\) is a perfect square trinomial.

Step 4: Solve the resulting equation

The perfect square trinomial can be written as \((x+3)^{2}=1\). Taking square root on both sides gives \(x+3=\pm\sqrt{1}\), hence we have two solutions \(x+3=1\) and \(x+3=-1\). Solving for \(x\) in both equations gives \(x=-2, -4\).